Calculation on fraction of square roots raised to the power 4

Question

Calculate $$\left(\frac{\sqrt{13}+\sqrt2-3}{\sqrt2}\right)^4+\left(\frac{\sqrt{13}-\sqrt2+3}{\sqrt2}\right)^4$$

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The numerator part can be re-written as $$\left(\frac{\sqrt{13}+(\sqrt2-3)}{\sqrt2}\right)^4+\left(\frac{\sqrt{13}-(\sqrt2-3)}{\sqrt2}\right)^4$$ It'd be nice if we can use the $(a+b)(a-b)=a^2-b^2$ (difference of squares) formula, which appeared in the numerator.

Answer 1

hint

Factor out by $ (\frac{\sqrt{13}}{\sqrt{2}})^4 $ to put it as

$$\frac{169}{4}\Bigl((1+a)^4+(1-a)^4\Bigr)$$

with $$a=\frac{\sqrt{2}-3}{\sqrt{13}},$$

$$(1+a)^4=1+4a+6a^2+4a^3+a^4,$$ $$(1-a)^4=1-4a+6a^2-4a^3+a^4,$$ $$13a^2=11-6\sqrt{2}\;,\;$$ and $$169a^4=193-132\sqrt{2}$$

Answer 2

Well, first of all we have:

$$\text{n}:=\left(\frac{\sqrt{13}+\sqrt{2}-3}{\sqrt{2}}\right)^4+\left(\frac{\sqrt{13}-\sqrt{2}+3}{\sqrt{2}}\right)^4\tag1$$

Let's rationalize the denominators and do some basic calculation:

\begin{equation} \begin{split} \text{n}&=\left(\frac{\sqrt{13}+\sqrt{2}-3}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\right)^4+\left(\frac{\sqrt{13}-\sqrt{2}+3}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\right)^4\\ \\ &=\left(\frac{\sqrt{2}\left(\sqrt{13}+\sqrt{2}-3\right)}{2}\right)^4+\left(\frac{\sqrt{2}\left(\sqrt{13}-\sqrt{2}+3\right)}{2}\right)^4\\ \\ &=\left(\frac{\sqrt{13}\sqrt{2}+\sqrt{2}\sqrt{2}-3\sqrt{2}}{2}\right)^4+\left(\frac{\sqrt{13}\sqrt{2}-\sqrt{2}\sqrt{2}+3\sqrt{2}}{2}\right)^4\\ \\ &=\left(\frac{\sqrt{26}-3\sqrt{2}+2}{2}\right)^4+\left(\frac{\sqrt{26}+3\sqrt{2}-2}{2}\right)^4\\ \\ &=\frac{\left(\sqrt{26}-3\sqrt{2}+2\right)^4}{2^4}+\frac{\left(\sqrt{26}+3\sqrt{2}-2\right)^4}{2^4}\\ \\ &=\frac{\left(\sqrt{26}-3\sqrt{2}+2\right)^4}{16}+\frac{\left(\sqrt{26}+3\sqrt{2}-2\right)^4}{16} \end{split}\tag2 \end{equation}

Now, let's write:

$$\left(\sqrt{26}-3\sqrt{2}+2\right)^4=\left(\underbrace{\left(\sqrt{26}-3\sqrt{2}+2\right)^2}_{:=\space\text{k}}\right)^2\tag3$$

So, for $\text{k}$ we get:

\begin{equation} \begin{split} \text{k}&=\left(\sqrt{26}-3\sqrt{2}+2\right)^2\\ \\ &=\left(\sqrt{26}-3\sqrt{2}+2\right)\left(\sqrt{26}-3\sqrt{2}+2\right)\\ \\ &=26-6\sqrt{13}+2\sqrt{26}+18-6\sqrt{2}-6\sqrt{13}+4-6\sqrt{2}+2\sqrt{26}\\ \\ &=4-6\sqrt{2}+2\sqrt{26} \end{split}\tag4 \end{equation}

And:

$$\text{k}^2=192-48 \sqrt{2}-48 \sqrt{13}+16 \sqrt{26}\tag5$$

I think you're able to finish now.

Answer 3

You can also calculate the value searched for using linear recurrences.

Set $$a = \frac{\sqrt{13}+\sqrt2-3}{\sqrt2}, b= \frac{\sqrt{13}-\sqrt2+3}{\sqrt2}$$

Then you have

$$a+b = \sqrt{26}, ab= 1+3\sqrt 2$$

Now, $a^4+b^4$ is $x_4$ in the linear recurrence $$x_{n+2} = \sqrt{26}x_{n+1}-(1+3\sqrt 2)x_n \text{ with }$$ $$x_0 = a^0+b^0 = 2, x_1 = a+b= \sqrt{26}$$

Just substituting and some simple calculations give

\begin{eqnarray*} x_2 & = &24 - 6\sqrt 2 \\ x_3 & = & \sqrt{26}(23 - 9\sqrt 2) \\ x_4 & = & 610 - 300\sqrt 2 \end{eqnarray*}